# Approximate non-invertible maps by invertible ones

Given a smooth map $$F:\mathbb{R}^n\to\mathbb{R}^n$$ with $$\det(DF)=0$$ is it possible to approximate it by smooth maps $$F_i:\mathbb{R}^n\to\mathbb{R}^n$$ with $$\det(DF_i)\neq 0$$ (maybe uniformly)?

• Is there any requirement for the approximation? For example, $n=2$. $F = (f(x,y), g(x,y))$, $\det (\mathrm{D}F) = 0$ means $f_x \cdot g_y = f_y \cdot g_x$. Apr 20, 2020 at 3:52
• I don't have any requirements for the approximation. This question was motivated by a book I'm reading where they say a proof can be reduced to the invertible case, without saying anything else.
– Sak
Apr 20, 2020 at 16:09
• By $\det DF=0$ do you mean $0$ everywhere?
– zhw.
Apr 22, 2020 at 16:37
• @zhw. If $\det(\mathrm{D}F)= 0$ means zero everywhere, for example, $n=2$, $F=(f(x,y), g(x,y))$ with $\det(\mathrm{D}F)= 0$, i.e., $\frac{\partial f}{\partial x} \cdot \frac{\partial g}{\partial y} = \frac{\partial f}{\partial y} \cdot \frac{\partial g}{\partial x}$ for all $(x,y)\in \mathbb{R}^2$, I guess, the question is: For any given $\epsilon > 0$, can we always find $F_1 = (f_1(x,y), g_1(x,y))$ such that $\det(\mathrm{D}F_1)\ne 0$ for any $(x,y)\in \mathbb{R}^2$, satisfying $|f_1(x,y) - f(x,y)| < \epsilon$ and $|g_1(x,y) - g(x,y)| < \epsilon$ for all $(x,y)\in \mathbb{R}^2$? Apr 26, 2020 at 4:13
If $$DF$$ is diagonalizable, an easy way could be to add $$\varepsilon(x_1,\dots,x_n)$$ and let $$\varepsilon\to 0$$. Otherwise, you should use the fact that nondiagonalizable matrices can be approximated by diagonalizable ones and adjust your $$F$$ appropriately to match the nondiagonalizable matrices. They have a nice explicit form: you can prove this by showing every matrix is similar to an upper triangular matrix.
• But $\epsilon$ depends on $x$, right? I agree that your method provides a map such that at one $x \in \mathbb{R}^n$ the jacobian is invertible. But can we do that for all $x \in \mathbb{R}^n$? Maybe I didn't understand really well the question..